S=ap^2+ap+c a quadratic question?

Question

if a quadratic equation S=ap^2+bp+c and, S=100 and P=29.95 on the first line, S=70 and P=39.95 on the second, and S=50 and P=49.95 on the third, how do I find the value of a, b and c? I have tried everything from matrices to elimination, and I still get the wrong answer, please help

Answer 1

You're trying to find the parabola that passes through the points (29.95, 100), (39.95, 70), and (49.95, 50). I'll take you through a standard solution, and finish up with a cute one (IMO).

The point (29.95, 100) tells us 100=a(29.95)^2 + b(29.95) + c, which is to say that
(1) 897.0025 a + 29.95 b + c = 100

Similarly, we get two more equations,
(2) 1596.0025 a + 39.95 b + c = 70 and
(3) 2495.0025 a + 49.95 b + c = 50.

Combining them (for the purposes of elimination):
(4): (2) - (1): 699a + 10 b = -30
(5): (3) - (2): 899a + 10 b = -20

(6): (5)-(4): 200a = 10 --> a=10/200=.05
Back to (4): 699(.05) + 10b = -30
10b = -30 - 34.95
10b = -64.95
b = -6.495

Back to (1): 897.0025(.05) + 29.95(-6.495) + c = 100
c = 100 + 29.95(6.495) - 897.0025(.05)
c = 100 + 194.52525 - 44.850125
c = 249.675125

This problem has a nice solution from Transformational Geometry, too. Notice how we start at (29.95, 100). We step 10 right and 30 down, to (39.95, 70). Next we step 10 right and 20 down, to (49.95, 50). Parabolas behave in such a way that the next step would be 10 right, 10 down--to (59.95, 40). Next we go 10 right and 0 down, to (69.95, 40).

This reminds me of the parabola y=x(x-1), which goes through such points as (-3,12), (-2,6), (-1,2), (0,0), (1,0). Reading y-coordinates left to right, they drop by 6, 4, 2, and 0--exactly one fifth the drop experienced by your parabola. For your parabola, we stepped right 10 each time; for mine, we step right in single units.

Long story short, we can transform my parabola into yours by:
(1) horizontally stretching it by a factor of 10
(2) vertically stretching it by a factor of 5
(3) moving the origin to (59.95, 40).

So, y = x(x-1) becomes:
(1) y = (x/10)((x/10)-1) = .01x(x-10)
(2) y = 5*.01x(x-10) = .05x(x-10)
(3) y-40 = .05(x-59.95)[(x-59.95)-10].

Simplifying the last equation, we get:
y= .05x^2 - 6.495 x + 249.675125. The same answer we got before.

Answer 2

Write three equations in three unknowns.

ap² + bp + c = S

a(29.95)² + b(29.95). + c = 100
a(39.95)² + b(39.95). + c = 70
a(49.95)² + b(49.95). + c = 50

Solving we get:

a = .05
b = -6.495
c = 249.675125

Answer 3

plug in the #s then use the three fomulas and use elimination... u should get the right answer that way